Subtract $ \dfrac{3}{1-x^3} $ from $ \dfrac{1}{1-x} $ to get $ \dfrac{ \color{purple}{ -x^2-x+2 } }{ x^3-1 }$.

To subtract raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the first fraction by $ \color{blue}{ -x^2-x-1 }$ and the second by $\color{blue}{ -1 }$.

$$ \begin{aligned} \frac{1}{1-x} - \frac{3}{1-x^3} & = \frac{ 1 \cdot \color{blue}{ \left( -x^2-x-1 \right) }}{ \left( 1-x \right) \cdot \color{blue}{ \left( -x^2-x-1 \right) }} - \frac{ 3 \cdot \color{blue}{ \left( -1 \right) }}{ \left( 1-x^3 \right) \cdot \color{blue}{ \left( -1 \right) }} = \\[1ex] &=\frac{ \color{purple}{ -x^2-x-1 } }{ -\cancel{x^2} -\cancel{x}-1+x^3+ \cancel{x^2}+ \cancel{x} } - \frac{ \color{purple}{ -3 } }{ -\cancel{x^2} -\cancel{x}-1+x^3+ \cancel{x^2}+ \cancel{x} } = \\[1ex] &=\frac{ \color{purple}{ -x^2-x+2 } }{ x^3-1 } \end{aligned} $$

Simplify $ \dfrac{-x^2-x+2}{x^3-1} $ to $ \dfrac{-x-2}{x^2+x+1} $.

Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x-1}$.

$$ \begin{aligned} \frac{-x^2-x+2}{x^3-1} & =\frac{ \left( -x-2 \right) \cdot \color{blue}{ \left( x-1 \right) }}{ \left( x^2+x+1 \right) \cdot \color{blue}{ \left( x-1 \right) }} = \\[1ex] &= \frac{-x-2}{x^2+x+1} \end{aligned} $$